$BD$ and $CE$ are bisectors of $\angle B$ and $\angle C$ of an isosceles $\triangle ABC$ with $AB = AC$. Prove that $BD = CE$.
Given:
$BD$ and $CE$ are bisectors of $\angle B$ and $\angle C$ of an isosceles $\triangle ABC$ with $AB = AC$.
To do:
We have to prove that $BD = CE$.
Solution:
In $\triangle ABC, AB = AC$
This implies,
$\angle B = \angle C$ (Angles opposite to equal sides are equal)
$\frac{1}{2} \angle B = \frac{1}{2} \angle C$
$\angle DBC = \angle ECB$
In $\triangle DBC$ and $\triangle EBC$,
$BC = BC$ (Common)
$\angle C = \angle B$
$\angle DBC = \angle ECB$
Therefore, by ASA axiom,
$\triangle DBC \cong \triangle EBC$
This implies,
$BD = CE$ (CPCT)
Hence proved.
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